Let $\displaystyle f\left ( x \right )=\left ( x-2 \right )\left ( x-3 \right )\left ( x-4 \right )\left ( x-5 \right )\left ( x-6 \right )$ then
- $\displaystyle f'\left ( x \right )=0$ has five real roots
- four roots of $\displaystyle f'\left ( x \right )=0$ lie in $\displaystyle \left ( 2, 3 \right )\cup \left ( 3, 4 \right )\cup \left ( 4, 5 \right )\cup \left ( 5, 6 \right )$
- the equation $\displaystyle f\left ( x \right )$ has only three roots
- four roots of $\displaystyle f\left ( x \right )= 0$ lie in $\displaystyle \left ( 1, 2 \right )\cup \left ( 2, 3 \right )\cup \left ( 3, 4 \right )\cup \left ( 4, 5 \right )$
By Rolle's Theorem, if f(x) has roots at 2, 3, 4, 5, 6, then f'(x) must have a root in each interval (2,3), (3,4), (4,5), and (5,6). This accounts for all 4 roots of the derivative of a degree 5 polynomial.
The function f(x) = (x-2)(x-3)(x-4)(x-5)(x-6) is a fifth-degree polynomial with five real, distinct roots at x = 2, 3, 4, 5, and 6. By Rolle's Theorem, between any two consecutive distinct real roots of a differentiable function, there must be at least one real root of its derivative, f'(x) = 0. Therefore, there is exactly one root of f'(x) in each of the intervals (2, 3), (3, 4), (4, 5), and (5, 6). Since f'(x) is a fourth-degree polynomial, these four roots account for all of its roots, meaning four roots of f'(x) = 0 lie in the union of those intervals.